$h(x)=2x\csc(x)\sec(x)$ Find $h'(x)$. Choose 1 answer: Choose 1 answer: (Choice A) A $2\csc(x)\sec(x)-2x\csc^2(x)+2x\sec^2(x)$ (Choice B) B $-2\csc(x)\cot(x)\sec(x)\tan(x)$ (Choice C) C $-2\csc(\sec(x))\cot(\sec(x))\sec(x)\tan(x)$ (Choice D) D $2x\sec^2(x)-2x\csc^2(x)$
Explanation: $2x\csc(x)\sec(x)$ is a product of three functions. Let... $u(x)=2x$ $v(x)=\csc(x)$ $w(x)=\sec(x)$... then $h(x)=u(x)\cdot v(x) \cdot w(x)$. To find $h'(x)$, we will need to use the product rule twice! $\begin{aligned} &\phantom{=}\dfrac{d}{dx}\left[u(x)\cdot v(x) \cdot w(x)\right] \\\\ &=u'(x)\cdot \Bigl(v(x)\cdot w(x) \Bigr)+u(x)\cdot\dfrac{d}{dx}\left[v(x)\cdot w(x)\right]&\gray{\text{Product rule}} \\\\ &=u'(x)\cdot \Bigl(v(x)\cdot w(x) \Bigr)+u(x)\cdot\Bigl(v'(x)\cdot w(x)+v(x)w'(x)\Bigr)&\gray{\text{Product rule}} \\\\ &= u'(x)\cdot v(x) \cdot w(x) + u(x) \cdot v'(x) \cdot w(x) + u(x) \cdot v(x) \cdot w'(x) \end{aligned}$ Let's differentiate $u$, $v$, and $w$ : $u'(x)=2$ $v'(x)=-\csc(x)\cot(x)$ $w'(x)=\sec(x)\tan(x)$ Now we can plug the equations for $u$, $v$, $w$, $u'$, $v'$, AND $w'$ into the expression we got: $\begin{aligned} &\phantom{=}u'(x)\cdot v(x) \cdot w(x) + u(x) \cdot v'(x) \cdot w(x) + u(x) \cdot v(x) \cdot w'(x) \\\\ &=2 \cdot \csc(x) \cdot \sec(x) + 2x \cdot -\csc(x)\cot(x) \cdot \sec(x) + 2x \cdot \csc(x) \cdot \sec(x)\tan(x) \\\\ &=2\csc(x)\sec(x)-2x\csc^2(x)+2x\sec^2(x) \end{aligned}$ In conclusion: $h'(x)=2\csc(x)\sec(x)-2x\csc^2(x)+2x\sec^2(x)$